I'm not sure if I understood correctly the question asked by OP, because the answer by Marco Ocram doesn't (in my opinion) answer what I understood of OP's questioning. So feel free to tell me if I am completely wrong with this answer and I will delete this one. I will use some maths but explained maths. If you don't have no time to read the whole thing, please skip to the end " To conclude"
From what I understood, anurag, your question asks for an explanation of why space-time is curved, and especially why time is dilated when a mass is here.
But first of all, I will point out that in general relativity the "gradient in time flow" does not create gravity. It is the energy content, momentum content, pressure content, shear stress content, and flow content of the object that creates gravity. And in a regime where all other effects than time dilation are negligible, time dilation is gravity in this context.
Now, back to your question. There are two possible answers. One saying 'this is the way nature works' and that is not satisfying at all. The other involves a few concepts of analytical mechanics and quantum mechanics à la Feynman:
There is something called the least action principle that 'can be justified' in point 2. This principle stated that the action functional must be extremized for the system characterized by it to reproduce classical mechanics. In classical mechanics, action can be thought of as the sum of a quantity called the lagrangian over time. At a given time $t$, the lagrangian gives us the difference between the kinetic energy and the potential energy of the studied system at this time. We then write:
\begin{equation}
\underbrace{S}_{\text{action}}=\underbrace{\int dt}_{\text{sum}} \,\,\underbrace{L}_{\text{Lagrangian}}=\int dt\left[\right. \underbrace{E_k}_{\text{Kin. Energy}}-\underbrace{V}_{\text{Pot. Energy}} \left. \right] \tag{1}
\end{equation}
Concretely the action can be thought of as the accumulation over time of the Lagrangian, that encodes the whole dynamics of the system.
In quantum mechanics there is something called 'Feynman path integrals' that is used to find the probability that the studied system goes from a state $A(t)$ at time $t$ to a state $B(t+\Delta t)$ in a given time interval $t+\Delta t$. Concretely it is the sum of a phase (that is representative of how much the action has a high value) over all the paths that can be chosen by the system beginning with the stated $A(t)$ and ending by the state $B(t+\Delta t)$. Here the word 'path' is to be understood as the configuration of the system (if it is a point then the configuration is its position, and if it is the whole space-time then its configuration is its curvature). We write it in different ways but one can find it formulated as follow:
\begin{equation}
\underbrace{\mathcal{A}(A(t)\rightarrow B(t+\Delta t))}_{\scriptstyle \text{Transition amplitude between} \atop \scriptstyle \text{the states $A(t)$ and $B(t+\Delta t)$}}=\underbrace{\int_A^B \mathcal{D} x}_{\scriptstyle \text{sum over the paths beginning} \atop \scriptstyle \text{at $A$ and ending at $B$}} \times \underbrace{e^{i S}}_{\scriptstyle \text{The phase giving how much} \atop \scriptstyle \text{the action has a high value}} \tag{2}
\end{equation}
In fact, the contribution of the paths that satisfy the least action principle is highly dominant in the transition amplitude when this very action has a value that is large in comparison to $\hbar$ (the reduced Planck constant, called 'h-bar')
Now time for space-time. The action for the space-time is written as:
\begin{equation}
S = \underbrace{\frac{1}{2\kappa}}_{\text{Coupling constant}}\times\int dt \underbrace{\iiint d^3 x \underbrace{\sqrt{-g} R}_{\text{Pondered curvature of space-time}}}_{\text{Lagrangian of space-time}} \tag{3}
\end{equation}
The lagrangian for space-time is roughly speaking the integral over the whole space of the curvature of space-time (just as I said before). If we add matter to this action, by simply adding the action of the considered matter, then the least action principle tells us this:
\begin{equation}
\underbrace{R_{\mu \nu}-\frac{1}{2}g_{\mu \nu} R}_{\text{Curvature of space-time}}=\underbrace{\kappa}_{\text{Coupling to matter}} \times \underbrace{T_{\mu \nu}}_{\text{Matter content of space-time}} \tag{4}
\end{equation}
That is, the known Einstein Field Equation (EFE).
Now, to 'justify' that this relation holds without involving the mysterious least action principle, I will do something really handwavy and I think I've seen this called 'Quantum gravity à la Hawking'. This goes as follow:
Define the space of configuration of space-time (really hard and unsolved problem)
Add matter to the system
Do the same thing as in (2) but with space-time
Search for the dominant paths contributions and it will match the least action principle provided that the total action of space-time + matter is large in comparison to $\hbar$ (at the classical level it is the case).
We write it symbolically like this :
\begin{equation}
\underbrace{\int \mathcal{D}g}_{\text{Sum of space-time config.}}\times \underbrace{\int \mathcal{D}\Phi}_{\text{Sum of matter config.}}\times \underbrace{e^{iS_\text{space-time}+iS_\text{matter}}}_{\text{Phase associated with space-time + matter}}
\tag{5}
\end{equation}
When considering only the configurations' paths beginning with a configuration $A$ of the Universe and ending to a configuration $B$ of the same Universe, we end up with a transition amplitude between $A$ and $B$.
To conclude: When time dilation is the only relevant effect of gravity, its cause is the fact that this very time dilation is just the Universe's behaving that contributes the most to the transition amplitude of 'Universe at time $t$' and 'Universe at time $t+\Delta t$'.
A bit more about 'at time $t$' for curved space-time itself:
In General Relativity we describe the curved space-time with coordinates that belong to a flat space-time. So when I say 'the transition amplitude of the Universe from time $t$ to time $t+\Delta t$', in fact, it is to be understood as the same but with 'for the infinitely far away observer' added.
As I said in the beginning I don't know if this is what you wanted as an answer, so tell me if I am wrong and I will delete this long post.
